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March 26th, 2018, 11:10 AM   #1
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Prove by mathematical induction

chelsea_lutchman@icloud.com



Prove by Mathematical Induction

n
sigma r^3 = n^2(n+1)^2/4
r = 1


Can someone show me how the do these sums? I've asked my lecturer for help but all he tells me is that it's too easy and he shouldn't have to help me. It would have been easier if he could at least give me an example about how it's done but he didn't. Google hasn't been a big help either. I'm not seeing any examples of induction with sigma that I can understand.

Last edited by notchelsea; March 26th, 2018 at 11:48 AM.
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March 26th, 2018, 12:50 PM   #2
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so far I have

1
sigma r^3 = 1^2(1+1)^2/2
r=1

1 = 1(4)/2

1 = 4/2

1 = 2

I'm not sure what to do after this for the k+1 case.

(took too long to edit so this is what I wanted to add)
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March 26th, 2018, 01:40 PM   #3
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Couple of things.

It is frequently a good idea to do the first few numbers by hand.

$\displaystyle \sum_{r=1}^1 r^3 = 1^3 = 1 = \dfrac{4}{4} = \dfrac{1 * 2^2}{4} = \dfrac{1^2(1 + 1)^2}{4}.$

Congratulations. You have already completed the first part of a proof by induction.

$\displaystyle \sum_{r=1}^2 r^3 = 1^3 + 2^3 = 9 = \dfrac{36}{4} = \dfrac{4 * 9}{4} = \dfrac{2^2(2 + 1)^2}{4}.$

$\displaystyle \sum_{r=1}^3 r^3 = 1^3 + 2^3 + 3^3 = 36 = \dfrac{4 * 36}{4} = \dfrac{9 * 16}{4} = \dfrac{3^2(3 + 1)^2}{4}.$

Doing this let's you understand what the theorem means, gives some confidence that it may well be true, and sometimes gives you a hint at a proof.

OK. The proof is in two stages.

$\text {PROVE: } \displaystyle n \in \mathbb N^+ \implies \sum_{r=1}^n r^3 = \dfrac{n^2(n + 1)^2}{4}.$

$n = 1 \implies \displaystyle \sum_{r=1}^n r^3 = 1^3 = 1 = \dfrac{1^2(1 + 1)^2}{4}.$

First stage is done.

$\therefore \exists \ k \in \mathbb N^+ \text { such that } k \ge 1 \text { and } \displaystyle \sum_{r= 1}^k r^3 = \dfrac{k^2(k + 1)^2}{4}.$

Now what you FREQUENTLY do is take the left hand side of the equation for k + 1 and try to break it down into the known expression in k and a simple expression in k + 1.

$\displaystyle \sum_{r=1}^{k+1}r^3 =$

$\displaystyle \left ( \sum_{r=1}^k r^3 \right ) + (k + 1)^3 =$

$\dfrac{k^2(k + 1)^2}{4} + k^3 + 3k^2 + 3k + 1 =$

$\dfrac{k^4 + 2k^3 + k^2}{4} + \dfrac{4k^3 + 12k^2 + 12k + 4}{4} =$

$\dfrac{k^4 + 6k^3 + 13k^2 + 12k + 4}{4} =$

$\dfrac{(k^4 + 4k^3 + 4k^2) + (2k^3 + 8k^2 + 8k) + (k^2 + 4k + 4)}{4} =$

$\dfrac{k^2(k^2 + 4k + 4) + 2k(k^2 + 4k + 4) + 1(k^2 + 4k + 4)}{4} =$

$\dfrac{(k^2 + 2k + 1)(k^2 + 4k + 4)}{4} = \dfrac{(k + 1)^2(k + 2)^2}{4} =$

$\dfrac{(k + 1)^2\{(k + 1) + 1\}^2}{4}.$

Done.

Last edited by JeffM1; March 26th, 2018 at 01:51 PM. Reason: Fixed missing parenthesis
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March 26th, 2018, 01:49 PM   #4
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Quote:
Originally Posted by notchelsea View Post
so far I have

1
sigma r^3 = 1^2(1+1)^2/2
r=1

1 = 1(4)/2

1 = 4/2

1 = 2

I'm not sure what to do after this for the k+1 case.

(took too long to edit so this is what I wanted to add)
Do you see that you messed up the case for n = 1. When you get an answer that
1 = 2, you can be fairly sure something came off the rails.
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